Magical seeds!

Let $N_j$ be the number of days until the $j$ th seed transforms. Then ,for any $1 \le j \le 6$ , $P[N_j = k] = 2^{-k}$ for all $k \ge 1$ , and so $P[N_j \le k] = 1 - 2^{-k}$ for all $k \ge 0$ . If $M$ is the number of days until all six seeds have transformed, then $M = \mathrm{max}(N_1,N_2,N_3,N_4,N_5,N_6)$ , and so $P[M \le k] \; = \; P[N_1,N_2,N_3,N_4,N_5,N_6 \le k] \; = \; P[N_1 \le k]^6 \; = \; \big(1 - 2^{-k}\big)^6$ for all $k \ge 0$ , and hence $P[m \ge k+1] \; = \; 1 - \big(1 - 2^{-k}\big)^6 \; = \; \sum_{j=1}^6 (-1)^{j-1}\binom{6}{j}2^{-jk}$ for all $k \ge 0$ , so that $E[M] \; = \; \sum_{k \ge 0}P[M \ge k+1] \; = \; \sum_{k=0}^\infty \sum_{j=1}^6 (-1)^{j-1}\binom{6}{j}2^{-jk} \; = \; \sum_{j=1}^6(-1)^{j-1}\binom{6}{j}\frac{2^j}{2^j-1} \; = \; \frac{7880}{1953}$ making the answer $7880 + 1953 = \boxed{9833}$ .